An $8$-cm-by-$8$-cm square is partitioned as shown. Points $A$ and $B$ are the midpoints of two opposite sides of the square. What is the area of the shaded region?

[asy]
draw((0,0)--(10,0));
draw((10,0)--(10,10));
draw((10,10)--(0,10));
draw((0,0)--(0,10));
draw((0,0)--(5,10));
draw((5,10)--(10,0));
draw((0,10)--(5,0));
draw((5,0)--(10,10));
fill((5,0)--(7.5,5)--(5,10)--(2.5,5)--cycle,gray);
label("A",(5,10),N);
label("B",(5,0),S);
[/asy]
Explanation: Draw a line segment from $A$ to $B,$ cutting the shaded diamond region in half. Next, draw the altitude from point $E$ to segment $AB.$ The new figure is shown below: [asy]
draw((0,0)--(10,0));
draw((10,0)--(10,10));
draw((10,10)--(0,10));
draw((0,0)--(0,10));
draw((0,0)--(5,10));
draw((5,10)--(10,0));
draw((0,10)--(5,0));
draw((5,0)--(10,10));
fill((5,0)--(7.5,5)--(5,10)--(2.5,5)--cycle,lightgray);
draw((5,0)--(5,10));
draw((5,5)--(7.5,5));
label("A",(5,10),N);
label("B",(5,0),S);
label("C",(10,0),S);
label("D",(10,10),N);
label("E",(7.5,5),E);
label("F",(5,5),W);
[/asy] $ABCD$ is a rectangle by symmetry of the square over line $AB.$ Thus, $\angle BAD = 90$ degrees. Since $\angle BAD = \angle BFE,$ we have $\triangle BFE \sim \triangle BAD.$ Since the diagonals of $ABCD$ bisect each other, $BE=BD/2,$ so the triangles are similar in a $1:2$ ratio. Thus, $FE$ is half the length of $AD,$ or $4/2=2$ cm.

The area of triangle $ABE$ is $$\frac{AB\cdot FE}{2}=\frac{8\cdot2}{2}=8.$$ The other half of the shaded region is identical and has the same area, so the entire shaded region has area $2\cdot8=\boxed{16}$ square cm.

We also might take a clever rearrangement approach. The two red pieces below can be rearranged to form a quadrilateral that is congruent to the gray quadrilateral, as can the two blue pieces, and as can the two green pieces. So, the area of the gray quadrilateral is $\frac 1 4$ of the area of the square. [asy]
fill((0,0)--(2.5,5)--(5,0)--cycle,red);
fill((0,10)--(2.5,5)--(5,10)--cycle,red);
fill((10,0)--(7.5,5)--(5,0)--cycle,green);
fill((10,10)--(7.5,5)--(5,10)--cycle,green);
fill((0,0)--(2.5,5)--(0,10)--cycle,blue);
fill((10,0)--(7.5,5)--(10,10)--cycle,blue);
draw((0,0)--(10,0));
draw((10,0)--(10,10));
draw((10,10)--(0,10));
draw((0,0)--(0,10));
draw((0,0)--(5,10));
draw((5,10)--(10,0));
draw((0,10)--(5,0));
draw((5,0)--(10,10));
fill((5,0)--(7.5,5)--(5,10)--(2.5,5)--cycle,gray);
label("A",(5,10),N);
label("B",(5,0),S);
[/asy]